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Chapter 9: Problem 23

Find the center, rertices, foci, and asymptotes of the hyperbola that satisfies the given equation, and sketch the hyperbola. $$y^{2}-(x+4)^{2}=1$$

Short Answer

Expert verified
The center is (-4,0), the vertices are (-4,卤1), the foci are (-4,卤\sqrt{2}), and the asymptotes are \(y = 卤x\).

Step by step solution

01

Identify the center

The center of a hyperbola is given by the -x and -y terms in the equation. Here, our center is (-4,0).
02

Identify the vertices

For a vertical hyperbola (which this is), the vertices are found at \(y 卤 a\), where \(a^2\) is the leading coefficient on the \(y^2\) term. Here, \(a = 1\), so our vertices are (-4, 卤1).
03

Identify the foci

The foci of a hyperbola are found at \(y 卤 \sqrt{a^2 + b^2}\), where \(b^2\) is the coefficient on the \(x^2\) term. Here, \(b = 1\), so our foci are (-4, 卤\sqrt{2}).
04

Identify the asymptotes

The asymptotes of a hyperbola are given by the equation \(y = 卤(a/b)x\). So here, our asymptotes are the lines \(y = 卤x\).
05

Sketch the Hyperbola

Now, with the center, vertices, and asymptotes identified, we can draw a rough sketch of the hyperbola. The center, vertices, and foci give us the dimensions of the hyperbola, and the asymptotes give us its shape and orientation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Center of Hyperbola
In a hyperbola, the center is an important reference point that helps in determining other key features such as the vertices and the asymptotes. For the given equation, \( y^{2} - (x+4)^{2} = 1 \), the form of the equation suggests it is centered not at the origin, but at a different point.

The standard form of a hyperbola is either \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \) for horizontal hyperbolas or \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \) for vertical hyperbolas. Here, the equation resembles the latter, indicating a vertical hyperbola.
  • Here, \( h = -4 \) and \( k = 0 \), so our center is \((-4, 0)\).
  • The center acts as the midpoint of the line segment joining the vertices and the foci.
Vertices of Hyperbola
Vertices define the most prominent points of a hyperbola, where it turns from one arm to the other. In the process of deriving them, it's necessary to know whether the hyperbola is vertical or horizontal, indicated by its equation.

For our vertical hyperbola equation \( y^{2} - (x+4)^{2} = 1 \), the vertices can be found using the center and the value of \( a \), which signifies half the distance between the vertices.
  • For the standard form \((y-k)^2/a^2 - (x-h)^2/b^2 = 1\), the vertices are at \((h, k \pm a)\).
  • Here, \(a^2 = 1\), giving \(a = 1\). This makes the vertices at \((-4, \pm 1)\).
Foci of Hyperbola
The foci of a hyperbola are two fixed points located along the transverse axis, beyond the vertices. These are crucial for understanding how the hyperbola is stretched and skewed.

For a vertical hyperbola like ours, given by \( y^{2} - (x+4)^{2} = 1 \), the foci are derived using the formula that combines both \( a^2 \) and \( b^2 \).
  • The foci are calculated with \((h, k \pm \sqrt{a^2 + b^2})\).
  • Given \( a^2 = 1 \) and \( b^2 = 1 \), \( a^2 + b^2 = 2 \).
  • Therefore, the foci are at \((-4, \pm \sqrt{2})\).
Asymptotes of Hyperbola
Asymptotes are lines that a hyperbola approaches as the branches extend infinitely. They aren't part of the hyperbola itself but help visualize its orientation and shape.

For the equation \( y^{2} - (x+4)^{2} = 1 \), the asymptotes of a vertical hyperbola can be identified using the formula \( y = \pm (a/b)x \). Since these lines act as invisible boundaries, they provide insights into how wide or narrow the hyperbola will be.
  • Here, note that \( a = b = 1 \), leading to \( y = \pm x \).
  • This establishes lines through the origin with slopes of +1 and -1, guiding the sketch of the hyperbola.

As you sketch, remember these asymptotes act as guidelines ensuring both arms of the hyperbola maintain their broader structure.

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Most popular questions from this chapter

Determine the equation in standard form of the hyperbola that satisfies the given conditions. Vertices at (5,-2),(1,-2)\(;\) slope of one asymptote is \(\frac{5}{2}\)

Determine the equations in standard form of two different hyperbolas that satisfy the given conditions. Center at (0,0)\(;\) transverse axis of length \(12 ;\) slope of one asymptote is 4

Graph each equation using a graphing utility. $$r=\frac{5.6}{1+0.7 \sin \theta}$$

An elliptical track is used for training race horses and their jockeys. Under normal circumstances, two coaches are stationed in the interior of the track, one at each focus, to observe the races and issue commands to the jockeys. One day, when only one coach was on duty, a horse threw its rider just as it reached the vertex closest to the vacant observation post. The coach at the other post called to the horse, which dutifully came running straight toward her. How far did the horse run before reaching the coach if the minor axis of the ellipse is 600 feet long and each observation post is 400 feet from the center of the interior of the track?

Determine the equations in standard form of two different ellipses that satisfy the given conditions. Major axis of length \(10 ;\) minor axis of length \(4 ;\) one vertex at (-5,-3)\(;\) major axis horizontal
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